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Does the following series have a closed-form expression:

$$\sum_{k=0}^{\infty} \frac{z^k}{(k!)^2}$$

I know that it must converge because:

$$\sum_{k=0}^{\infty} \frac{z^k}{k!} = e^z$$

and the $(k!)^2$ denominator obviously increases more quickly than the $k!$ denominator.

This problem came up in computing the probability of a draw in a football match with each team's goal scoring modeled as a Poisson process.

Thanks,

John

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  • $\begingroup$ Will let someone else derive it if they'd like but wolfram says it's related to bessel I wolframalpha.com/input/… $\endgroup$ – spaceisdarkgreen Oct 11 '17 at 0:14
  • $\begingroup$ the bessel functions appear to be in the realm of $\sum\frac{z^2}{k!^2}$ $\endgroup$ – Dan Uznanski Oct 11 '17 at 0:34
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Denote

$$f(z)=\sum_{n=0}^\infty\frac{z^n}{(n!)^2}$$

By term-wise differentiation, we find that

$$f'(z)+zf''(z)=f(z)$$

A rather simple differential equation with the general solution

$$f(z)=c_1I_0(2\sqrt z)+c_2K_0(2\sqrt z)$$

where $I_n$ is a modified Bessel function of the first kind, and $K_n$ is a modified Bessel function of the second kind. By using

$$f(0)=f'(0)=1$$

we find that

$$f(z)=I_0(2\sqrt z)$$

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